Consider 125 mL of 0.110 M MgCl2 solution.
a. How many moles of MgCl2 are contained in 125 mL of 0.0110 M MgCl2?
b. What is the concentration of Mg2+ ion?
c. What is the concentration of Cl- ion?
moles MgCl2 = M x L = ??
b. Since there is 1 atom Mg per molecule of MgCl2, the concn Mg = concn MgCl2.
c. Since there are two atoms Cl per molecule of MgCl2, the concn of Cl ions is twice that of MgCl2 or Mg.
To find the answers to these questions, we will use the equation:
Moles = concentration (in M) x volume (in L)
First, let's convert the given volume from milliliters (mL) to liters (L):
125 mL = 125/1000 = 0.125 L
a. To find the number of moles of MgCl2, we will use the concentration and the volume:
Number of moles of MgCl2 = 0.0110 M x 0.125 L = 0.00138 moles
Therefore, there are 0.00138 moles of MgCl2 in 125 mL of 0.0110 M MgCl2 solution.
b. To find the concentration of Mg2+ ion, we need to consider that MgCl2 ionizes into one mole of Mg2+ and two moles of Cl- ions.
Since the molar ratio is 1:1, the concentration of Mg2+ ions will be the same as the concentration of MgCl2.
The concentration of Mg2+ ion is 0.0110 M.
c. Similarly, since the molar ratio of MgCl2 to Cl- is 1:2, the concentration of Cl- ions will be twice the concentration of MgCl2.
Concentration of Cl- ion = 2 x 0.0110 M = 0.0220 M
Therefore, the concentration of Cl- ion is 0.0220 M.